Question

prepare questions on two coins, three coins tossed in probability​

Answer 1

Step-by-step explanation:

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting 3 heads = {HHH}

\therefore n(E) = 1

\therefore P(E) = \frac{n(E)}{n(S)} = \frac{1}{8}

The required probability of getting 3 heads is \frac{1}{8}.